An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-6.3,0)\) and \((6.3,0)\) and a covertex at \((0, 6)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (4.7, 5.05):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(6.3)^2 + (6)^2 =a^2\]
\[a ~=~ \sqrt{6.3^2+6^2} ~=~ 8.7\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,6) to \(F_2\)\[2a = 17.4\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=6.3\]\[2c = 12.6\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(4.7-(-6.3))^2+5.05^2}\]
\[s=12.1038217\]\[s\approx12.1\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(4.7-(6.3))^2+5.05^2}\]
\[w=5.297405\]\[w\approx5.3\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=17.4\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-4,0)\) and \((4,0)\) and a covertex at \((0, 4.2)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (4.21, 2.89):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(4)^2 + (4.2)^2 =a^2\]
\[a ~=~ \sqrt{4^2+4.2^2} ~=~ 5.8\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,4.2) to \(F_2\)\[2a = 11.6\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=4\]\[2c = 8\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(4.21-(-4))^2+2.89^2}\]
\[s=8.7038038\]\[s\approx8.7\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(4.21-(4))^2+2.89^2}\]
\[w=2.8976197\]\[w\approx2.9\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=11.6\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-3.6,0)\) and \((3.6,0)\) and a covertex at \((0, 7.7)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (7.31, 3.93):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(3.6)^2 + (7.7)^2 =a^2\]
\[a ~=~ \sqrt{3.6^2+7.7^2} ~=~ 8.5\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,7.7) to \(F_2\)\[2a = 17\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=3.6\]\[2c = 7.2\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(7.31-(-3.6))^2+3.93^2}\]
\[s=11.5962494\]\[s\approx11.6\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(7.31-(3.6))^2+3.93^2}\]
\[w=5.4045351\]\[w\approx5.4\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=17\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-1.1,0)\) and \((1.1,0)\) and a covertex at \((0, 6)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (5.53, 2.53):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(1.1)^2 + (6)^2 =a^2\]
\[a ~=~ \sqrt{1.1^2+6^2} ~=~ 6.1\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,6) to \(F_2\)\[2a = 12.2\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=1.1\]\[2c = 2.2\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(5.53-(-1.1))^2+2.53^2}\]
\[s=7.096323\]\[s\approx7.1\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(5.53-(1.1))^2+2.53^2}\]
\[w=5.1015488\]\[w\approx5.1\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=12.2\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-4.2,0)\) and \((4.2,0)\) and a covertex at \((0, 4)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (3.17, 3.35):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(4.2)^2 + (4)^2 =a^2\]
\[a ~=~ \sqrt{4.2^2+4^2} ~=~ 5.8\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,4) to \(F_2\)\[2a = 11.6\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=4.2\]\[2c = 8.4\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(3.17-(-4.2))^2+3.35^2}\]
\[s=8.0956408\]\[s\approx8.1\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(3.17-(4.2))^2+3.35^2}\]
\[w=3.5047682\]\[w\approx3.5\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=11.6\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-2.8,0)\) and \((2.8,0)\) and a covertex at \((0, 4.5)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (3.6, 3.3):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(2.8)^2 + (4.5)^2 =a^2\]
\[a ~=~ \sqrt{2.8^2+4.5^2} ~=~ 5.3\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,4.5) to \(F_2\)\[2a = 10.6\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=2.8\]\[2c = 5.6\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(3.6-(-2.8))^2+3.3^2}\]
\[s=7.2006944\]\[s\approx7.2\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(3.6-(2.8))^2+3.3^2}\]
\[w=3.3955854\]\[w\approx3.4\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=10.6\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-2.1,0)\) and \((2.1,0)\) and a covertex at \((0, 7.2)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (4.3, 5.9):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(2.1)^2 + (7.2)^2 =a^2\]
\[a ~=~ \sqrt{2.1^2+7.2^2} ~=~ 7.5\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,7.2) to \(F_2\)\[2a = 15\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=2.1\]\[2c = 4.2\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(4.3-(-2.1))^2+5.9^2}\]
\[s=8.7045965\]\[s\approx8.7\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(4.3-(2.1))^2+5.9^2}\]
\[w=6.2968246\]\[w\approx6.3\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=15\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-4.5,0)\) and \((4.5,0)\) and a covertex at \((0, 2.8)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (4.59, 1.4):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(4.5)^2 + (2.8)^2 =a^2\]
\[a ~=~ \sqrt{4.5^2+2.8^2} ~=~ 5.3\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,2.8) to \(F_2\)\[2a = 10.6\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=4.5\]\[2c = 9\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(4.59-(-4.5))^2+1.4^2}\]
\[s=9.1971789\]\[s\approx9.2\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(4.59-(4.5))^2+1.4^2}\]
\[w=1.4028899\]\[w\approx1.4\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=10.6\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-6.5,0)\) and \((6.5,0)\) and a covertex at \((0, 7.2)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (8.96, 2.76):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(6.5)^2 + (7.2)^2 =a^2\]
\[a ~=~ \sqrt{6.5^2+7.2^2} ~=~ 9.7\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,7.2) to \(F_2\)\[2a = 19.4\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=6.5\]\[2c = 13\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(8.96-(-6.5))^2+2.76^2}\]
\[s=15.7044325\]\[s\approx15.7\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(8.96-(6.5))^2+2.76^2}\]
\[w=3.6971881\]\[w\approx3.7\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=19.4\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
An ellipse is a set of points such that the total distance from (1) one focus to (2) the edge to (3) the other focus is constant for any point on the edge. The two foci (\(F_1\) and \(F_2\)) are a distance \(2c\) apart, and the total distance from \(F_1\) to the edge to \(F_2\) is a distance \(2a\).
Let’s consider the example below with foci at \((-2.8,0)\) and \((2.8,0)\) and a covertex at \((0, 9.6)\):
Evaluate \(2a\) and \(2c\).
\(a=\)
\(2a=\)
\(c=\)
\(2c=\)
Now, let’s pick another point on the ellipse, with (approximate) coordinates (4.99, 8.32):
The distance \(s\) is from \(F_1\) to the point on the edge. The distance \(w\) is from \(F_2\) to the point on the edge. Evaluate \(s\), \(w\), and the total distance (\(s+w\)).
\(s=\)
\(w=\)
\(s+w=\)
Remember, the total distance when connecting
The first focus \(F_1\)
Any point on the edge
The second focus \(F_2\)
equals \(2a\). Based on this information, find the location of the right vertex at (\(x\), 0) in the diagram below.
\(x=\)
Solution
To find \(a\), use the Pythagorean Theorem. The length \(a\) is a hypotenuse of a right triangle with legs along the axes.
\[(2.8)^2 + (9.6)^2 =a^2\]
\[a ~=~ \sqrt{2.8^2+9.6^2} ~=~ 10\]
Use multiplication by 2 to find the total distance from \(F_1\) to (0,9.6) to \(F_2\)\[2a = 20\]
It is easy to find \(c\) from the graph. It is the distance from the center to either focus. In this graph, the center and the focus are on the same horizontal axis, so distance is just a difference, and the subtrahend is 0.
\[c=2.8\]\[2c = 5.6\]
Use pythagorean theorem to find \(s\).
\[s = \sqrt{(4.99-(-2.8))^2+8.32^2}\]
\[s=11.3976533\]\[s\approx11.4\]
Use pythagorean theorem to find \(w\).
\[w = \sqrt{(4.99-(2.8))^2+8.32^2}\]
\[w=8.6034005\]\[w\approx8.6\]
Add \(s\) and \(w\) to get the total distance.
\[s+w=20\]
Notice that \(s+w=2a\), because the total distance is always \(2a\) for any point on the edge.
To find \(x\), use the fact that the total distance is \(2a\).
So:
\[\begin{align}
h &= -2\\\\
k &= -5\\\\
r_1 &= 7\\\\
r_2 &= 3
\end{align}\]
Substitute those numbers into the equation:
\[\frac{(x+2)^2}{49}+\frac{(y+5)^2}{9}=1\]
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 16 meters apart, and she uses a string that is 20 meters long (ignoring the amount needed to tie around the stake).
How long is the major axis of the generated ellipse?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=20\).
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 4.8 meters apart, and she uses a string that is 14.8 meters long (ignoring the amount needed to tie around the stake).
How long is the major axis of the generated ellipse?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=14.8\).
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 11.2 meters apart, and she uses a string that is 13 meters long (ignoring the amount needed to tie around the stake).
How long is the major axis of the generated ellipse?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=13\).
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 3.2 meters apart, and she uses a string that is 13 meters long (ignoring the amount needed to tie around the stake).
How long is the major axis of the generated ellipse?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=13\).
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 2.6 meters apart, and she uses a string that is 17 meters long (ignoring the amount needed to tie around the stake).
How long is the major axis of the generated ellipse?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=17\).
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 14.4 meters apart, and she uses a string that is 18 meters long (ignoring the amount needed to tie around the stake).
How long is the major axis of the generated ellipse?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=18\).
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 2.6 meters apart, and she uses a string that is 17 meters long (ignoring the amount needed to tie around the stake).
How long is the major axis of the generated ellipse?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=17\).
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 13 meters apart, and she uses a string that is 19.4 meters long (ignoring the amount needed to tie around the stake).
How long is the major axis of the generated ellipse?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=19.4\).
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 6.6 meters apart, and she uses a string that is 13 meters long (ignoring the amount needed to tie around the stake).
How long is the major axis of the generated ellipse?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=13\).
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 11 meters apart, and she uses a string that is 14.6 meters long (ignoring the amount needed to tie around the stake).
How long is the major axis of the generated ellipse?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=14.6\).
Question
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 9.6 meters apart, and she uses a string that is 14.6 meters long (ignoring the amount needed to tie around the stake).
How long is the minor axis of the generated ellipse?
Solution
The string is 14.6 meters long; divide by 2 to find \(a\).
\[a = \frac{14.6}{2} = 7.3\]
The stakes are 9.6 meters apart; divide by 2 to find \(c\).
\[c = \frac{9.6}{2} = 4.8\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[4.8^2+b^2=7.3^2 \]
\[b^2=7.3^2-4.8^2\]
\[b^2=53.29-23.04\]
\[b^2=30.25\]
\[b=\sqrt{30.25}\]
\[b=5.5\]
The length of the minor axis is twice \(b\).
\[2b=11\]
I guess I should also mention, that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 3.2 meters apart, and she uses a string that is 13 meters long (ignoring the amount needed to tie around the stake).
How long is the minor axis of the generated ellipse?
Solution
The string is 13 meters long; divide by 2 to find \(a\).
\[a = \frac{13}{2} = 6.5\]
The stakes are 3.2 meters apart; divide by 2 to find \(c\).
\[c = \frac{3.2}{2} = 1.6\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[1.6^2+b^2=6.5^2 \]
\[b^2=6.5^2-1.6^2\]
\[b^2=42.25-2.56\]
\[b^2=39.69\]
\[b=\sqrt{39.69}\]
\[b=6.3\]
The length of the minor axis is twice \(b\).
\[2b=12.6\]
I guess I should also mention, that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 2.2 meters apart, and she uses a string that is 12.2 meters long (ignoring the amount needed to tie around the stake).
How long is the minor axis of the generated ellipse?
Solution
The string is 12.2 meters long; divide by 2 to find \(a\).
\[a = \frac{12.2}{2} = 6.1\]
The stakes are 2.2 meters apart; divide by 2 to find \(c\).
\[c = \frac{2.2}{2} = 1.1\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[1.1^2+b^2=6.1^2 \]
\[b^2=6.1^2-1.1^2\]
\[b^2=37.21-1.21\]
\[b^2=36\]
\[b=\sqrt{36}\]
\[b=6\]
The length of the minor axis is twice \(b\).
\[2b=12\]
I guess I should also mention, that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 13 meters apart, and she uses a string that is 19.4 meters long (ignoring the amount needed to tie around the stake).
How long is the minor axis of the generated ellipse?
Solution
The string is 19.4 meters long; divide by 2 to find \(a\).
\[a = \frac{19.4}{2} = 9.7\]
The stakes are 13 meters apart; divide by 2 to find \(c\).
\[c = \frac{13}{2} = 6.5\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[6.5^2+b^2=9.7^2 \]
\[b^2=9.7^2-6.5^2\]
\[b^2=94.09-42.25\]
\[b^2=51.84\]
\[b=\sqrt{51.84}\]
\[b=7.2\]
The length of the minor axis is twice \(b\).
\[2b=14.4\]
I guess I should also mention, that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 10.8 meters apart, and she uses a string that is 18 meters long (ignoring the amount needed to tie around the stake).
How long is the minor axis of the generated ellipse?
Solution
The string is 18 meters long; divide by 2 to find \(a\).
\[a = \frac{18}{2} = 9\]
The stakes are 10.8 meters apart; divide by 2 to find \(c\).
\[c = \frac{10.8}{2} = 5.4\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[5.4^2+b^2=9^2 \]
\[b^2=9^2-5.4^2\]
\[b^2=81-29.16\]
\[b^2=51.84\]
\[b=\sqrt{51.84}\]
\[b=7.2\]
The length of the minor axis is twice \(b\).
\[2b=14.4\]
I guess I should also mention, that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 12.8 meters apart, and she uses a string that is 16 meters long (ignoring the amount needed to tie around the stake).
How long is the minor axis of the generated ellipse?
Solution
The string is 16 meters long; divide by 2 to find \(a\).
\[a = \frac{16}{2} = 8\]
The stakes are 12.8 meters apart; divide by 2 to find \(c\).
\[c = \frac{12.8}{2} = 6.4\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[6.4^2+b^2=8^2 \]
\[b^2=8^2-6.4^2\]
\[b^2=64-40.96\]
\[b^2=23.04\]
\[b=\sqrt{23.04}\]
\[b=4.8\]
The length of the minor axis is twice \(b\).
\[2b=9.6\]
I guess I should also mention, that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 5.6 meters apart, and she uses a string that is 10.6 meters long (ignoring the amount needed to tie around the stake).
How long is the minor axis of the generated ellipse?
Solution
The string is 10.6 meters long; divide by 2 to find \(a\).
\[a = \frac{10.6}{2} = 5.3\]
The stakes are 5.6 meters apart; divide by 2 to find \(c\).
\[c = \frac{5.6}{2} = 2.8\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[2.8^2+b^2=5.3^2 \]
\[b^2=5.3^2-2.8^2\]
\[b^2=28.09-7.84\]
\[b^2=20.25\]
\[b=\sqrt{20.25}\]
\[b=4.5\]
The length of the minor axis is twice \(b\).
\[2b=9\]
I guess I should also mention, that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 7.8 meters apart, and she uses a string that is 17.8 meters long (ignoring the amount needed to tie around the stake).
How long is the minor axis of the generated ellipse?
Solution
The string is 17.8 meters long; divide by 2 to find \(a\).
\[a = \frac{17.8}{2} = 8.9\]
The stakes are 7.8 meters apart; divide by 2 to find \(c\).
\[c = \frac{7.8}{2} = 3.9\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[3.9^2+b^2=8.9^2 \]
\[b^2=8.9^2-3.9^2\]
\[b^2=79.21-15.21\]
\[b^2=64\]
\[b=\sqrt{64}\]
\[b=8\]
The length of the minor axis is twice \(b\).
\[2b=16\]
I guess I should also mention, that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 9.6 meters apart, and she uses a string that is 14.6 meters long (ignoring the amount needed to tie around the stake).
How long is the minor axis of the generated ellipse?
Solution
The string is 14.6 meters long; divide by 2 to find \(a\).
\[a = \frac{14.6}{2} = 7.3\]
The stakes are 9.6 meters apart; divide by 2 to find \(c\).
\[c = \frac{9.6}{2} = 4.8\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[4.8^2+b^2=7.3^2 \]
\[b^2=7.3^2-4.8^2\]
\[b^2=53.29-23.04\]
\[b^2=30.25\]
\[b=\sqrt{30.25}\]
\[b=5.5\]
The length of the minor axis is twice \(b\).
\[2b=11\]
I guess I should also mention, that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She places the two stakes (pins) 12.8 meters apart, and she uses a string that is 16 meters long (ignoring the amount needed to tie around the stake).
How long is the minor axis of the generated ellipse?
Solution
The string is 16 meters long; divide by 2 to find \(a\).
\[a = \frac{16}{2} = 8\]
The stakes are 12.8 meters apart; divide by 2 to find \(c\).
\[c = \frac{12.8}{2} = 6.4\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[6.4^2+b^2=8^2 \]
\[b^2=8^2-6.4^2\]
\[b^2=64-40.96\]
\[b^2=23.04\]
\[b=\sqrt{23.04}\]
\[b=4.8\]
The length of the minor axis is twice \(b\).
\[2b=9.6\]
I guess I should also mention, that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 17 meters and the length of the minor axis to be 16.8 meters.
How long should the string be?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=17\).
Question
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 10.6 meters and the length of the minor axis to be 9 meters.
How long should the string be?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=10.6\).
Question
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 14.6 meters and the length of the minor axis to be 9.6 meters.
How long should the string be?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=14.6\).
Question
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 19.4 meters and the length of the minor axis to be 14.4 meters.
How long should the string be?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=19.4\).
Question
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 10.6 meters and the length of the minor axis to be 5.6 meters.
How long should the string be?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=10.6\).
Question
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 17.8 meters and the length of the minor axis to be 16 meters.
How long should the string be?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=17.8\).
Question
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 17 meters and the length of the minor axis to be 15.4 meters.
How long should the string be?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=17\).
Question
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 17 meters and the length of the minor axis to be 15.4 meters.
How long should the string be?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=17\).
Question
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 13 meters and the length of the minor axis to be 12.6 meters.
How long should the string be?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=13\).
Question
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 17 meters and the length of the minor axis to be 15 meters.
How long should the string be?
Solution
Surprisingly, the length of the major axis exactly matches the length of the string. Both of them are a length \(2a=17\).
Question
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 17.8 meters and the length of the minor axis to be 7.8 meters.
How far apart should the stakes be placed?
Solution
If the major axis is 17.8 meters long, then the string is 17.8 meters long; divide by 2 to find \(a\).
\[a = \frac{17.8}{2} = 8.9\]
The length minor axis is 7.8 meters. Divide this by two to find the length of the semi-minor axis (the shorter radius).
\[b = \frac{7.8}{2} = 3.9\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[3.9^2+c^2=8.9^2 \]
\[c^2=8.9^2-3.9^2\]
\[c^2=79.21-15.21\]
\[c^2=64\]
\[c=\sqrt{64}\]
\[c=8\]
The distance between the stakes is \(2c\).
\[2c=16\]
I guess I should also mention, that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 19.4 meters and the length of the minor axis to be 13 meters.
How far apart should the stakes be placed?
Solution
If the major axis is 19.4 meters long, then the string is 19.4 meters long; divide by 2 to find \(a\).
\[a = \frac{19.4}{2} = 9.7\]
The length minor axis is 13 meters. Divide this by two to find the length of the semi-minor axis (the shorter radius).
\[b = \frac{13}{2} = 6.5\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[6.5^2+c^2=9.7^2 \]
\[c^2=9.7^2-6.5^2\]
\[c^2=94.09-42.25\]
\[c^2=51.84\]
\[c=\sqrt{51.84}\]
\[c=7.2\]
The distance between the stakes is \(2c\).
\[2c=14.4\]
I guess I should also mention, that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 10.6 meters and the length of the minor axis to be 9 meters.
How far apart should the stakes be placed?
Solution
If the major axis is 10.6 meters long, then the string is 10.6 meters long; divide by 2 to find \(a\).
\[a = \frac{10.6}{2} = 5.3\]
The length minor axis is 9 meters. Divide this by two to find the length of the semi-minor axis (the shorter radius).
\[b = \frac{9}{2} = 4.5\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[4.5^2+c^2=5.3^2 \]
\[c^2=5.3^2-4.5^2\]
\[c^2=28.09-20.25\]
\[c^2=7.84\]
\[c=\sqrt{7.84}\]
\[c=2.8\]
The distance between the stakes is \(2c\).
\[2c=5.6\]
I guess I should also mention, that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 13 meters and the length of the minor axis to be 11.2 meters.
How far apart should the stakes be placed?
Solution
If the major axis is 13 meters long, then the string is 13 meters long; divide by 2 to find \(a\).
\[a = \frac{13}{2} = 6.5\]
The length minor axis is 11.2 meters. Divide this by two to find the length of the semi-minor axis (the shorter radius).
\[b = \frac{11.2}{2} = 5.6\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[5.6^2+c^2=6.5^2 \]
\[c^2=6.5^2-5.6^2\]
\[c^2=42.25-31.36\]
\[c^2=10.89\]
\[c=\sqrt{10.89}\]
\[c=3.3\]
The distance between the stakes is \(2c\).
\[2c=6.6\]
I guess I should also mention, that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 10.6 meters and the length of the minor axis to be 5.6 meters.
How far apart should the stakes be placed?
Solution
If the major axis is 10.6 meters long, then the string is 10.6 meters long; divide by 2 to find \(a\).
\[a = \frac{10.6}{2} = 5.3\]
The length minor axis is 5.6 meters. Divide this by two to find the length of the semi-minor axis (the shorter radius).
\[b = \frac{5.6}{2} = 2.8\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[2.8^2+c^2=5.3^2 \]
\[c^2=5.3^2-2.8^2\]
\[c^2=28.09-7.84\]
\[c^2=20.25\]
\[c=\sqrt{20.25}\]
\[c=4.5\]
The distance between the stakes is \(2c\).
\[2c=9\]
I guess I should also mention, that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 13 meters and the length of the minor axis to be 6.6 meters.
How far apart should the stakes be placed?
Solution
If the major axis is 13 meters long, then the string is 13 meters long; divide by 2 to find \(a\).
\[a = \frac{13}{2} = 6.5\]
The length minor axis is 6.6 meters. Divide this by two to find the length of the semi-minor axis (the shorter radius).
\[b = \frac{6.6}{2} = 3.3\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[3.3^2+c^2=6.5^2 \]
\[c^2=6.5^2-3.3^2\]
\[c^2=42.25-10.89\]
\[c^2=31.36\]
\[c=\sqrt{31.36}\]
\[c=5.6\]
The distance between the stakes is \(2c\).
\[2c=11.2\]
I guess I should also mention, that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 13 meters and the length of the minor axis to be 12.6 meters.
How far apart should the stakes be placed?
Solution
If the major axis is 13 meters long, then the string is 13 meters long; divide by 2 to find \(a\).
\[a = \frac{13}{2} = 6.5\]
The length minor axis is 12.6 meters. Divide this by two to find the length of the semi-minor axis (the shorter radius).
\[b = \frac{12.6}{2} = 6.3\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[6.3^2+c^2=6.5^2 \]
\[c^2=6.5^2-6.3^2\]
\[c^2=42.25-39.69\]
\[c^2=2.56\]
\[c=\sqrt{2.56}\]
\[c=1.6\]
The distance between the stakes is \(2c\).
\[2c=3.2\]
I guess I should also mention, that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 17.8 meters and the length of the minor axis to be 16 meters.
How far apart should the stakes be placed?
Solution
If the major axis is 17.8 meters long, then the string is 17.8 meters long; divide by 2 to find \(a\).
\[a = \frac{17.8}{2} = 8.9\]
The length minor axis is 16 meters. Divide this by two to find the length of the semi-minor axis (the shorter radius).
\[b = \frac{16}{2} = 8\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[8^2+c^2=8.9^2 \]
\[c^2=8.9^2-8^2\]
\[c^2=79.21-64\]
\[c^2=15.21\]
\[c=\sqrt{15.21}\]
\[c=3.9\]
The distance between the stakes is \(2c\).
\[2c=7.8\]
I guess I should also mention, that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 17 meters and the length of the minor axis to be 16.8 meters.
How far apart should the stakes be placed?
Solution
If the major axis is 17 meters long, then the string is 17 meters long; divide by 2 to find \(a\).
\[a = \frac{17}{2} = 8.5\]
The length minor axis is 16.8 meters. Divide this by two to find the length of the semi-minor axis (the shorter radius).
\[b = \frac{16.8}{2} = 8.4\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[8.4^2+c^2=8.5^2 \]
\[c^2=8.5^2-8.4^2\]
\[c^2=72.25-70.56\]
\[c^2=1.69\]
\[c=\sqrt{1.69}\]
\[c=1.3\]
The distance between the stakes is \(2c\).
\[2c=2.6\]
I guess I should also mention, that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
A gardener is making an ellipse using the pins and string method. She wants the length of the major axis to be 19 meters and the length of the minor axis to be 15.2 meters.
How far apart should the stakes be placed?
Solution
If the major axis is 19 meters long, then the string is 19 meters long; divide by 2 to find \(a\).
\[a = \frac{19}{2} = 9.5\]
The length minor axis is 15.2 meters. Divide this by two to find the length of the semi-minor axis (the shorter radius).
\[b = \frac{15.2}{2} = 7.6\]
Draw a diagram
Notice the right triangle. We know two sides, and we are hoping to find the third side. Use Pythagorean Equation.
\[7.6^2+c^2=9.5^2 \]
\[c^2=9.5^2-7.6^2\]
\[c^2=90.25-57.76\]
\[c^2=32.49\]
\[c=\sqrt{32.49}\]
\[c=5.7\]
The distance between the stakes is \(2c\).
\[2c=11.4\]
I guess I should also mention, that if \(x^2+y^2=z^2\), then \((2x)^2+(2y)^2=(2z)^2\). This implies we can also get our answer without halving and doubling.
